Why KMnO₄ Matters
Manganese sits at the centre of the first transition series and displays the widest range of oxidation states of any 3d metal, reaching the group oxidation number of +7. That maximum state is never found as a simple cation; it is stabilised only inside oxoanions, where small, highly charged manganese is wrapped in oxide ions. Potassium permanganate, $\ce{KMnO4}$, is the everyday embodiment of Mn(VII), and NCERT pairs it with potassium dichromate as the two industrially and analytically important oxoanion salts of the d-block.
The deep purple of permanganate, its violent oxidising power, and its sensitivity to the pH of the medium make it a recurring NEET subject. The compound also offers a clean teaching example of charge-transfer colour, of d-electron counting, and of how electrode potentials and stoichiometry are governed by the surrounding solution. Mastering it means mastering a cluster of ideas, not a single fact.
| Quantity | Manganate | Permanganate |
|---|---|---|
| Formula | MnO4^2- | MnO4^- |
| Mn oxidation state | +6 | +7 |
| d-electron count | d1 | d0 |
| Colour | Dark green | Deep purple |
| Magnetism | Paramagnetic (1 unpaired e⁻) | Diamagnetic |
| Geometry | Tetrahedral | Tetrahedral |
Preparation from Pyrolusite
The starting ore is pyrolusite, $\ce{MnO2}$, in which manganese is in the +4 state. Permanganate requires Mn(VII), so the synthesis is necessarily a two-stage climb in oxidation number: first up to +6 as green manganate, then up to +7 as purple permanganate. No single fusion step delivers permanganate cleanly, which is why the NCERT description always separates the two conversions.
Stage 1 — Fusion to manganate, MnO₄²⁻
Pyrolusite is fused with an alkali metal hydroxide together with an oxidising agent such as $\ce{KNO3}$, or simply with air. The melt turns into a dark green mass of potassium manganate, in which manganese has reached the +6 state:
$$\ce{2MnO2 + 4KOH + O2 -> 2K2MnO4 + 2H2O}$$
The same oxidative fusion can be written with $\ce{KNO3}$ acting as the oxidant in place of atmospheric oxygen. The dark green solid is potassium manganate, $\ce{K2MnO4}$, which is stable in strongly alkaline solution or in pure water.
Stage 2 — Oxidation to permanganate, MnO₄⁻
The manganate is then raised to permanganate by any of three routes. The most distinctive is disproportionation: in a neutral or acidic solution the +6 manganate is simultaneously oxidised to +7 permanganate and reduced to +4 manganese dioxide. Even a trace of acid is enough to trigger it.
$$\ce{3MnO4^2- + 4H+ -> 2MnO4^- + MnO2 + 2H2O}$$
Commercially the preferred route is the alkaline oxidative fusion of $\ce{MnO2}$ followed by electrolytic oxidation of manganate(VI) at the anode, which converts all the manganese to permanganate rather than wasting one-third of it as $\ce{MnO2}$:
$$\ce{MnO4^2- -> MnO4^- + e-} \quad \text{(anodic oxidation)}$$
In the laboratory a manganese(II) salt can be taken straight to permanganate using the powerful oxidant peroxodisulphate:
$$\ce{2Mn^2+ + 5S2O8^2- + 8H2O -> 2MnO4^- + 10SO4^2- + 16H+}$$
In the disproportionation of manganate, how is the +6 manganese distributed between the two products?
Of three $\ce{MnO4^2-}$ ions, two are oxidised from +6 to +7 (forming $\ce{MnO4^-}$) and one is reduced from +6 to +4 (forming $\ce{MnO2}$). The electrons released by the two oxidised ions are exactly the electrons absorbed by the one reduced ion, so the process is self-contained — the hallmark of a disproportionation.
Structure of Manganate & Permanganate
Both the manganate and permanganate ions are tetrahedral, with the four oxygen atoms arranged symmetrically around the central manganese. The bonding is not purely sigma: the source notes that significant pπ–dπ bonding occurs through overlap of the filled p orbitals of oxygen with the d orbitals of manganese. This multiple-bond character shortens the Mn–O distances and helps stabilise the very high oxidation states that would otherwise be untenable for a simple cation.
Four equivalent O atoms at the corners of a tetrahedron; Mn(VII) in $\ce{MnO4^-}$ is d0, Mn(VI) in $\ce{MnO4^2-}$ is d1. The geometry is identical for both ions; only the d-electron count differs.
$\ce{KMnO4}$ itself forms dark purple, almost black crystals that are isostructural with potassium perchlorate, $\ce{KClO4}$ — a useful structural analogy, since the perchlorate ion is likewise a symmetric tetrahedral oxoanion of a central atom in its highest oxidation state.
Colour, Charge Transfer & Magnetism
This is the section that separates careful candidates from the rest. The colour and magnetism of the two ions follow directly from the d-electron count of manganese, and they behave in opposite ways.
Manganate, $\ce{MnO4^2-}$, has Mn in +6, which is a d1 configuration. The single unpaired electron makes the ion paramagnetic, and that same lone d electron can undergo a d–d transition, which is responsible for the green colour.
Permanganate, $\ce{MnO4^-}$, has Mn in +7, a d0 configuration. With no d electrons at all there can be no unpaired electron, so the ion is diamagnetic, and there can be no d–d transition. Yet permanganate is one of the most intensely coloured ions known. The colour must therefore come from somewhere else: a ligand-to-metal charge-transfer (LMCT) transition, in which an electron jumps from an oxygen-based orbital into an empty manganese d orbital. Charge-transfer bands are fully allowed and so are far more intense than the weak, formally forbidden d–d bands, which is why a few crystals of $\ce{KMnO4}$ colour a large volume of water deep purple.
Permanganate colour is NOT a d–d transition
A standard distractor states that the purple of $\ce{MnO4^-}$ arises from a d–d transition. It cannot — Mn(VII) is d0 and has no d electron to promote. The colour is a charge-transfer band. Only the green manganate (d1) shows a genuine d–d transition.
d0 ($\ce{MnO4^-}$) → charge transfer, diamagnetic. d1 ($\ce{MnO4^2-}$) → d–d transition, paramagnetic.
Dichromate is the other great d-block oxidant. See how its preparation and acidic-medium reactions differ in Potassium Dichromate (K₂Cr₂O₇).
Medium-Dependent Oxidation
Permanganate is a powerful oxidant, but the product it is reduced to — and therefore the number of electrons it accepts — depends sharply on the acidity of the solution. NCERT states the point plainly: the hydrogen-ion concentration plays an important part in influencing the reaction. Three distinct half-reactions cover the three regimes.
| Medium | Reduction half-reaction | Product (Mn state) | e⁻ gained | E° / V |
|---|---|---|---|---|
| Acidic | MnO4^- + 8H+ + 5e- -> Mn^2+ + 4H2O |
Mn²⁺ (+2), colourless | 5 | +1.52 |
| Neutral / faintly alkaline | MnO4^- + 4H+ + 3e- -> MnO2 + 2H2O |
MnO₂ (+4), brown | 3 | +1.69 |
| Strongly alkaline | MnO4^- + e- -> MnO4^2- |
MnO₄²⁻ (+6), green | 1 | +0.56 |
The three half-reactions, written explicitly:
$$\ce{MnO4^- + 8H+ + 5e- -> Mn^2+ + 4H2O} \qquad (E^\circ = +1.52\ \text{V})$$
$$\ce{MnO4^- + 4H+ + 3e- -> MnO2 + 2H2O} \qquad (E^\circ = +1.69\ \text{V})$$
$$\ce{MnO4^- + e- -> MnO4^2-} \qquad (E^\circ = +0.56\ \text{V})$$
Same oxidant, three products: the medium dictates whether $\ce{MnO4^-}$ accepts 5, 3 or 1 electron. The acidic route to $\ce{Mn^2+}$ is the basis of nearly all permanganate titrations.
A subtlety worth noting is that thermodynamics is not the whole story. At $[\ce{H+}] = 1$, permanganate should oxidise water, but in practice the reaction is extremely slow unless manganese(II) ions are present or the temperature is raised. The kinetics of permanganate reactions therefore matter as much as the electrode potentials, which is why warming and autocatalysis appear repeatedly in its titrations.
Match the product to the medium, not the reverse
Examiners give a balanced equation and ask which medium it represents — or give the medium and ask for the product. Anchor on the n-factor: acidic = 5e⁻ → Mn²⁺, neutral/faintly alkaline = 3e⁻ → MnO₂, strongly alkaline = 1e⁻ → MnO₄²⁻. The acidic route is the strongest oxidant in practice and the only one used quantitatively.
Key Titration Reactions
Almost all analytical use of permanganate exploits the acidic half-reaction. Two molecules of $\ce{MnO4^-}$ supply the equivalent of five oxygen atoms, so the balanced equations always show a coefficient of 2 on permanganate against a coefficient of 5 on the species supplying two electrons. The three NEET-favourite reductants are oxalate, iron(II) and iodide.
Oxidation of oxalate (and oxalic acid)
Oxalate is oxidised to carbon dioxide; the reaction is slow at room temperature and is carried out warm, around 333 K, after which liberated $\ce{Mn^2+}$ autocatalyses it:
$$\ce{5C2O4^2- + 2MnO4^- + 16H+ -> 2Mn^2+ + 8H2O + 10CO2}$$
Oxidation of iron(II)
Pale green iron(II) is oxidised to yellow iron(III), the basis of permanganometric estimation of iron:
$$\ce{5Fe^2+ + MnO4^- + 8H+ -> Mn^2+ + 4H2O + 5Fe^3+}$$
Oxidation of iodide
In acidic medium iodide is oxidised all the way to iodine:
$$\ce{10I^- + 2MnO4^- + 16H+ -> 2Mn^2+ + 8H2O + 5I2}$$
The medium changes the iodide outcome dramatically. In neutral or faintly alkaline solution, where $\ce{MnO2}$ is the manganese product, iodide is instead oxidised further to iodate:
$$\ce{2MnO4^- + H2O + I^- -> 2MnO2 + 2OH^- + IO3^-}$$
| Reductant | Acidic-medium product | Diagnostic colour change |
|---|---|---|
| Oxalate / oxalic acid | CO2 | Purple → colourless (warm, autocatalysed) |
| Iron(II), Fe²⁺ | Fe^3+ | Pale green → yellow |
| Iodide, I⁻ | I2 | Purple discharged, iodine liberated |
| Nitrite, NO₂⁻ | NO3^- | Purple discharged |
| Hydrogen sulphide / sulphide | S (precipitate) | Yellow turbidity of sulphur |
| Sulphite / sulphurous acid | SO4^2- | Purple discharged |
For completeness, two further acidic-medium half-equations from the source are the oxidation of sulphite and of nitrite:
$$\ce{5SO3^2- + 2MnO4^- + 6H+ -> 2Mn^2+ + 3H2O + 5SO4^2-}$$
$$\ce{5NO2^- + 2MnO4^- + 6H+ -> 2Mn^2+ + 5NO3^- + 3H2O}$$
Never acidify a permanganate titration with HCl
Permanganate is strong enough to oxidise chloride to chlorine, so hydrochloric acid would consume permanganate independently of the analyte and inflate the titre. The acid of choice is dilute sulphuric acid, because sulphate is not oxidised. This is a frequent one-line statement question.
Acidify with dilute H₂SO₄, never HCl — HCl gives a high, unreliable reading.
Physical Properties & Uses
Potassium permanganate forms dark purple, almost black crystals that are sparingly soluble in water (about 6.4 g per 100 g of water at 293 K), giving the familiar deep purple solution. Beyond its colour, it shows diamagnetism with a weak, temperature-dependent paramagnetism — a fine point that NCERT attributes to molecular-orbital effects beyond the syllabus. On heating to about 513 K the solid decomposes, evolving oxygen:
$$\ce{2KMnO4 ->[\Delta] K2MnO4 + MnO2 + O2}$$
The applications all trace back to its oxidising power. In analytical chemistry it is the reagent of permanganometry. In preparative organic chemistry it is a favourite oxidant, cleaving and oxidising a range of functional groups. Industrially it is used for bleaching wool, cotton, silk and other textile fibres, and for the decolourisation of oils — uses that depend directly on its strong oxidising action.
Potassium permanganate in one screen
- Source & route: from pyrolusite $\ce{MnO2}$ → fusion with KOH + oxidant → green $\ce{K2MnO4}$ (Mn +6) → oxidation/disproportionation → purple $\ce{KMnO4}$ (Mn +7).
- Disproportionation: $\ce{3MnO4^2- + 4H+ -> 2MnO4^- + MnO2 + 2H2O}$.
- Structure: both ions tetrahedral with pπ–dπ bonding; $\ce{KMnO4}$ isostructural with $\ce{KClO4}$.
- Colour/magnetism: $\ce{MnO4^2-}$ (d¹) green, paramagnetic, d–d transition; $\ce{MnO4^-}$ (d⁰) purple, diamagnetic, charge transfer.
- Medium rule: acidic → Mn²⁺ (5e⁻, 1.52 V); neutral → MnO₂ (3e⁻, 1.69 V); strongly alkaline → MnO₄²⁻ (1e⁻, 0.56 V).
- Titration: acidify with dilute H₂SO₄ (never HCl); oxalate warmed to 333 K, Fe²⁺→Fe³⁺, I⁻→I₂.